an extension theorem for a sequence of krein space...

Research ArticleAn Extension Theorem for a Sequence of KreinSpace Contractions

GeraldWanjala

Department of Mathematics and Statistics Sultan Qaboos University PO Box 36 Al-Khod 123 Muscat Oman

Correspondence should be addressed to Gerald Wanjala wanjalagyahoocom

Received 8 December 2017 Revised 23 January 2018 Accepted 30 January 2018 Published 1 March 2018

Academic Editor Gelu Popescu

Copyright copy 2018 Gerald Wanjala This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Consider Krein spaces U and Y and let H119896 and K119896 be regular subspaces of U and Y respectively such that H119896 sub H119896+1 andK119896 sub K119896+1 (119896 isin N) For each 119896 isin N let 119860119896 H119896 rarr K119896 be a contraction We derive necessary and sufficient conditions for theexistence of a contraction 119861 Urarr Y such that 119861|H119896 = 119860119896 Some interesting results are proved along the way

1 Introduction

A number of extensions and generalizations of classical func-tion theoretic interpolation problems are driven by factorssuch as applications in systems and control theoryMany suchextensions and generalizations make use of the commutantlifting theorem in one way or another In particular thecommutant lifting theorem in the Hilbert space case whichwas obtained by Sz-Nagy and Foias has been used tosolve extension problems like the ones of Nevanlinna-PickNudelman Nehari and many others Extensions of thistheorem to an indefinite setting are given in [1ndash5] In [6](see also [7 8]) a time-variant version of the commutantlifting theorem is developed This time-variant version iscalled the three-chain completion theorem and is used to solvea number of nonstationary norm constrained interpolationproblems on Hilbert spaces Recall (see [6]) that the givendata for the three-chain completion problem are boundedlinear operators119860119896 H119896 997888rarrK119896 ⊖M119896 (119896 isin Z) (1)

where H119896 sub U and M119896 sub K119896 sub Y for 119896 isin Z are Hilbertspaces satisfying the inclusion relations

H119896minus1 subH119896M119896minus1 subM119896

K119896minus1 subK119896 (119896 isin Z) (2)

Given operators (1) and tolerance 120574 the problem is to find anoperator 119861 Urarr Y such that 119861 le 120574 and119861H119896 subK119896(119868 minus 119875M119896) 11986110038161003816100381610038161003816H119896 = 119860119896 (119896 isin Z) (3)

As far as we know no extension of this theorem to anindefinite setting has been developed so far

The three-chain theorem mentioned above is the moti-vation for the extension problem considered in this paperFor 119896 isin N consider a sequence of Krein space contractions119860119896 H119896 rarr K119896 where H119896 and K119896 are nested regularsubspaces of some Krein spaces U and Y respectively Theproblem is to find a contraction 119861 Urarr Y such that119861ℎ = 119860119896ℎ (4)

for all ℎ isinH119896In order to keep this paper as self-contained as possible

we briefly outline some definitions and some elementary factsabout Krein spaces and bounded linear operators defined

HindawiInternational Journal of Mathematics and Mathematical SciencesVolume 2018 Article ID 5178454 7 pageshttpsdoiorg10115520185178454

2 International Journal of Mathematics and Mathematical Sciences

on them This is done in Section 2 Some important resultsare stated and proved in Section 3 An extension theorem isconsidered in Section 4 while Section 5 contains some simpleapplication of the extension theorem discussed in Section 4

2 Preliminaries

In this section we recall some definitions and basic notions ofindefinite metric space theory More details can be obtainedfrom [2 9ndash12]

LetK be a linear space and let ⟨sdot sdot⟩ be a sesquilinear formonKThis sesquilinear form is called an indefinite metric onK IfK admits a direct orthogonal sum decomposition

K =K+ oplusKminus (5)

in which (Kplusmn plusmn⟨sdot sdot⟩) are Hilbert spaces then K (or(K ⟨sdot sdot⟩)) is called aKrein space Decomposition (5) which isnot unique in general is called a fundamental decompositionand gives rise to orthogonal projections from K onto Kplusmnwhich we denote by 119875plusmn respectively The self-adjoint andunitary operator 119869 on K defined by 119869 = 119875+ minus 119875minus is calleda fundamental symmetry onK

The spaceK with the inner product[119909 119910] = ⟨119869119909 119910⟩ = ⟨119909+ 119910+⟩ minus ⟨119909minus 119910minus⟩ 119909plusmn 119910plusmn isinKplusmn (6)

is a Hilbert space This inner product is used to define thenorm of an element 119909 of the Krein spaceK by1199092 = [119909 119909] (7)

Topological notions such as convergence and continuity areunderstood to be with respect to this norm topology Theinner product [sdot sdot] in (6) depends on decomposition (5) asdoes the norm sdot in (7) but the norms generated by differentdecompositions ofK are equivalentWewill denote the norm sdot in (7) by sdot 119869 where clarity is needed

An orthogonal projection in a Krein space is a boundedself-adjoint operator 119877 inK such that 1198772 = 119877 Note that thenorm of an orthogonal projection in a Krein space need notbe less than 1 The rangeM of an orthogonal projection 119877 ina Krein space K is a closed subspace of K and the space Kcan be decomposed as

K =M oplusMperp (8)

where Mperp = 119897 isin K ⟨119897 119898⟩ = 0 forall119898 isin MOn the other hand given a closed subspace M of Ksuch that decomposition (8) holds then M is the range ofan orthogonal projection in K In this case (M ⟨sdot sdot⟩) isagain a Krein space Unlike in the Hilbert space case thedecomposition (8) need not hold for a given closed subspaceClosed subspaces for which this decomposition holds arereferred to as regular subspaces

Let H and K be Krein spaces and let 119879 H rarr K be abounded linear operator We say that 119879 is a contraction if forall 119909 isinH ⟨119879119909 119879119909⟩K le ⟨119909 119909⟩H If both119879 and its Krein spaceadjoint 119879lowast are contractions then 119879 is called a bicontraction

Let H and K be Krein spaces and let 119879 isin 119861(HK) thespace of bounded linear operators fromH intoK A columnextension of 119879 is an operator of the form

119862 = (119879119864) H 997888rarr (KE) (9)

whereE is aKrein space and119864 isin 119861(HE) By a row extensionof 119879 we shall mean an operator of the form

119871 = (119879 119864) (HE) 997888rarrK (10)

where E is a Krein space and 119864 isin 119861(EK) It is shownin [2] (see also [3]) that if 119879 is a contraction then thereexist contractive row and column extensions of 119879 where theextension space is a Hilbert space Contractive 2 times 2 matrixextensions of a contractive operator 119879 H rarr K where theextending spaceE is a Hilbert space are thoroughly discussedin [2]where Lemma 1Theorem2 andLemma3 can be foundResults more general than those provided by Lemma 1 andTheorem 2 can be found in [1] while Lemma 3 can also befound in [4]

Lemma 1 LetH andK be Krein spaces and let119879 isin 119861(HK)be a contraction Let

119862 = ( 119864lowast) isin 119861 (HK oplus E) (11)

be a contractive column extension of 119879 with E a Hilbert spaceIf norms are computed relative to some fixed fundamentaldecompositions of H and K and the induced fundamentaldecomposition ofK oplus E then1198622 le 1 + 2 1198792 (12)

The above norm estimate enables one to fix a bound forthe norm of the operator 119862 The following lemma is helpfulin finding 2times2matrix extensions of a contractive operator 119879See [13] for a similar result in a Hilbert space setting

Theorem 2 LetHK and E be Krein spaces and let E be aHilbert space Assume that

(11986211 11986212) (HE) 997888rarrK

(1198621111986221) H 997888rarr (KE) (13)

are contractions Then there exists an operator 11986222 E rarr Esuch that

119862 = (11986211 1198621211986221 11986222) (HE) 997888rarr (KE) (14)

is a contraction If (11986211 11986212) (15)

is a bicontraction11986222may be chosen such that119862 is a bicontrac-tion

International Journal of Mathematics and Mathematical Sciences 3

We conclude this section by stating the following lemma

Lemma 3 Let N1N2 be regular subspaces of a Kreinspace H such that N1 sup N2 sup sdot sdot sdot Suppose that theprojections 1198751 1198752 of H onto the subspaces N1N2 areuniformly bounded Then

N = infin⋂119899=1

N119899M = spanNperp119899 119899 = 1 2

(16)

are regular subspaces ofHwithN =Mperp If 119875 is the projectionof H onto N then 119875 = lim119899rarrinfin119875119899 with convergence in thestrong operator topology

3 Some Preliminary Results

In this section we prove some useful results regardingsequences of regular subspaces See [4 Corollary 33 andLemma 34] for closely related results

Theorem 4 For 119899 ge 1 let L119899 be a sequence of regularsubspaces of a Krein space H such that L119899 sub L119899+1 119899 isin NFor each 119899 isin N let 119875119899 be the projection of H onto L119899 If theprojections 119875119899 are uniformly bounded then L = spanL119899 is aregular subspace ofH

Proof Set N1 = Lperp1 N2 = Lperp2 Then N1N2 is asequence of regular subspaces of H such that N1 sup N2 supsdot sdot sdot Let1198761 1198762 be the projections ofH onto the subspacesN1N2 Then 119876119899 = (1 minus 119875119899) and so 119876119899 le 1 + 119875119899Since the projections 119875119899 are uniformly bounded we see thatthe projections119876119899 are also uniformly bounded HenceL is aregular subspace ofH by Lemma 3

Theorem 5 For 119899 ge 1 letL119899 be regular subspaces of a Kreinspace K such that L119899 sub L119899+1 Let 119875119899 be the projection of Konto L119899 If L119899 = L+119899 oplusLminus119899 is a fundamental decompositionofL119899 we denote by 119876plusmn119899 the projection ofL119899 ontoLplusmn119899 Let

L = spanL119899 (17)

Then the following are equivalent

(1) L is regular and there exists a fundamental symmetry119869 on L such that 119869 | L119899 is a fundamental symmetryonL119899

(2) There exist fundamental decompositions L119899 = L+119899 oplusLminus119899 such that

(a) Lplusmn119899 subLplusmn119899+1(b) sup119876plusmn119899119875119899119869 lt infin

where 119869 is any fundamental symmetry onK

Proof Suppose that (1) holds and let L = L+ oplus Lminus bethe fundamental decomposition ofL which gives rise to the

fundamental symmetry 119869 = 119876+minus119876minus with the stated propertywhere 119876plusmnL = Lplusmn Then 119869 | L119899 = 119876+119899 minus 119876minus119899 where 119876plusmn119899L119899 =Lplusmn119899 To show that (a) holds we let 119909 isinL+119899 Then119909 = 119876+119899119909 minus 119876minus119899119909 = 119869119909 = 119876+119899+1119909 minus 119876minus119899+1119909 (18)

Hence 119909 = 119876+119899+1119909minus119876minus119899+1119909 and so (119868 minus119876+119899+1)119909 = minus119876minus119899+1119909 Thismeans that 119876minus119899+1119909 = minus119876minus119899+1119909 and that 119876minus119899+1119909 = 0 Therefore119909 = 119876+119899+1119909 isin L+119899+1 Similarly we have that Lminus119899 sub Lminus119899+1Hence (a) holds To show that (b) also holds we let119910 isinK =L119899 oplus (L ⊖L119899) oplusLperp (19)

Let 119869119899 = 119869 |L119899 119869119899 = 119869 | (L⊖L119899) and let 119869perp be a fundamentalsymmetry onLperpThen 119869 fl 119869+119869perp is a fundamental symmetryonK Now100381710038171003817100381711991010038171003817100381710038172K119869 = ⟨119869119910 119910⟩= ⟨119869 (119909119899 + 119910119899 + 119911119899) (119909119899 + 119910119899 + 119911119899)⟩

= ⟨119869119899119909119899 119909119899⟩ + ⟨119869119899119910119899 119910119899⟩ + ⟨119869perp119911119899 119911119899⟩ (20)

where 119910 = 119909119899 + 119910119899 + 119911119899 with 119909119899 isin L119899 119910119899 isin (L ⊖L119899) and119911119899 isinLperp Therefore1003817100381710038171003817119876+11989911987511989911991010038171003817100381710038172L119899119869119899 = 1003817100381710038171003817119876+11989911990911989910038171003817100381710038172L119899 119869119899 = ⟨119869119899119876+119899119909119899 119876+119899119909119899⟩= ⟨119876+119899119909119899 119876+119899119909119899⟩ = ⟨119876+119899119909119899 119909119899⟩= ⟨(119876+119899 minus 119876minus119899 ) 119909119899 119909119899⟩ + ⟨119876minus119899119909119899 119909119899⟩le ⟨(119876+119899 minus 119876minus119899 ) 119909119899 119909119899⟩ = ⟨119869119899119909119899 119909119899⟩le ⟨119869119899119909119899 119909119899⟩ + ⟨119869119899119910119899 119910119899⟩ + ⟨119869perp119911119899 119911119899⟩= 100381710038171003817100381711991010038171003817100381710038172K

(21)

Hence 119876+119899119875119899 le 1 Similarly 119876minus119899119875119899 le 1Conversely assume that (2) holds We start by showing

thatL is regular To do this we note that 119875119899 = (119876+119899 + 119876minus119899 )119875119899So 119875119899 le 119876+119899119875119899 + 119876minus119899119875119899 le 1198721 for some constant1198721 gt 0HenceL = spanL119899 is regular byTheorem 4

Next set N119899 = (L+119899 )perp and let 119877119899 be the projectionof K onto N119899 = (L+119899 )perp Then 119877119899 = 119868 minus 119876+119899119875119899 Hence119877119899 le 1 + 119876+119899119875119899 le 1198722 for some constant 1198722 gt 0since (2) holds Hence N+ fl ⋂N119899 and M+ fl spanL+119899are regular subspaces by Lemma 3 Let 119909 isin M+ Thenthere exists a sequence 119909119899 isin spanL+119899 such that 119909119899 rarr 119909Hence ⟨119909119899 119909119899⟩ rarr ⟨119909 119909⟩ This means that ⟨119909 119909⟩ ge 0 since⟨119909119899 119909119899⟩ ge 0 Hence M+ is regular and nonnegative and soit is a uniformly positive subspace of L Similarly Mminus flspanLminus119899 is a uniformly negative subspace ofLWenow showthat

L =M+ oplusMminus (22)


Let 119909 isin M+ Then there exists a sequence 119909119899 isin span L+119899such that 119909119899 rarr 119909 If 119910 isin Mminus then there exists a sequence119910119899 isin span Lminus119899 such that 119910119899 rarr 119910 Hence ⟨119909 119910⟩ = lim⟨119909119899 119910119899⟩= 0 This implies that M+ perp Mminus Next assume that 119906 isinL ⊖ (M+ oplusMminus) Then 119906 perp M+ and 119906 perp Mminus This in turnmeans that 119906 perp L+119899 and 119906 perp Lminus119899 and so 119906 perp L119899 Let V isin LThen there exists a sequence V119899 isin spanL119899 such that V119899 rarr VHence ⟨119906 V⟩ = lim⟨119906 V119899⟩ = 0 for all V isin L and so V = 0Hence L = M+ oplus Mminus This implies that M+ is maximalpositive andMminus ismaximal negativeHenceM+ is amaximaluniformly positive space and Mminus is a maximal uniformlynegative space Since M+ perp Mminus M+ is uniformly positiveand Mminus is uniformly negative there exists a fundamentaldecomposition L = L+ oplus Lminus such that Mplusmn sub Lplusmn ButMplusmn maximal implies Mplusmn = Lplusmn Hence L = M+ oplus Mminus

is a fundamental decomposition of L This decompositiongives rise to a fundamental symmetry 119869 = 119876+ minus 119876minus where119876plusmnL =Mplusmn

We show that 119869 | L119899 = 119876+119899 minus 119876minus119899 Let 119909 isin L119899 sub LThen 119909 = 119909+ + 119909minus 119909plusmn isin Lplusmn119899 sub spanLplusmn119899 = Mplusmn Hence119876+119909 = 119876+(119909+ + 119909minus) = 119909+ = 119876+119899119909 Similarly 119876minus119909 = 119876minus119899119909 andso 119869 |L119899 = 119876+119899 minus 119876minus119899 Corollary 6 For 119899 ge 1 let K L119899 L and 119875119899 be as inTheorem 5 Then there exists a fundamental symmetry 119869 onKwhich commutes with the projections 119875119899 ontoL119899 if and only ifL is regular and there exists a fundamental symmetry 1198691015840 onLsuch that 1198691015840 |L119899 is a fundamental symmetry onL119899

Proof Suppose that L is regular and 1198691015840 exists which has thestated property Let L = L+ oplus Lminus be the fundamentaldecomposition which gives rise to 1198691015840 SinceL is regular thereexists a fundamental decomposition K = K+ oplus Kminus suchthatLplusmn subKplusmn This gives rise to 119869 onK such that 119869 |L119899 is afundamental symmetry onL119899 Let119891 isinKThen119891 = 119891119899+119891perp119899 where 119891119899 isin L119899 and 119891perp119899 isin Lperp119899 Now 119869119875119899119891 = 119869119891119899 = 119891+119899 minus 119891minus119899(where 119891plusmn119899 isinLplusmn119899 L119899 =L+119899 oplusLminus119899 ) On the other hand

119875119899119869119891 = 119875119899119869 (119891119899 + 119891perp119899 )= 119875119899119869 [(119891+119899 + 119891minus119899 ) + (119891perp+119899 + 119891perpminus119899 )]= 119875119899 [119869 (119891+119899 + 119891minus119899 ) + 119869 (119891perp+119899 + 119891perpminus119899 )]= 119875119899 (119891+119899 minus 119891minus119899 ) + 119875119899 (119891perp+119899 minus 119891perpminus119899 ) = 119891+119899 minus 119891minus119899 (23)

Hence 119869119875119899 = 119875119899119869Conversely assume that there exists a fundamental sym-

metry 119869onK that commuteswith the projections119875119899 To showthat119875119899rsquos are uniformly boundedwe consider the fundamentaldecomposition K = K+ oplus Kminus which gives rise to thefundamental symmetry 119869 Let

119875119899 = (11986011989911 1198601198991211986011989921 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (24)

be the matrix representation of 119875119899 with respect to thesedecompositions Then the commutativity condition

(1 00 minus1)(11986011989911 1198601198991211986011989921 11986011989922) = (119860

11989911 1198601198991211986011989921 11986011989922)(1 00 minus1) (25)

implies that 11986011989921 = 11986011989912 = 0 and so the matrix representationof 119875119899 is diagonal that is to say

119875119899 = (11986011989911 00 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (26)

Since 11986011989911 le 1 and 11986011989922 le 1 for all 119899 isin N we see that119875119899rsquos are uniformly boundedThe uniformboundedness of119875119899rsquosimplies that L is regular From the matrix representation of119875119899 above we see that 11986011989911 = 119876+119899119875119899 and 11986011989922 = 119876minus119899119875119899 and sosup119876plusmn119899119875119899 lt infin Hence condition (b) in Theorem 5 holdsSince for each 119899 isin N L119899 is a regular subspace and L119899 subL119899+1 we have that for any fundamental decompositionL119899 =L+119899 oplusLminus119899 there exist a fundamental decomposition

L119899+1 =L+119899+1 oplusLminus119899+1 (27)

such thatLplusmn119899 subLplusmn119899+1 Hence condition (a) inTheorem 5 alsoholds This shows that part (1) of Theorem 5 holds and thiscompletes the proof

4 An Extension Theorem fora Sequence of Contractions

In this section we formulate and give a proof of an extensiontheorem for a sequence of contractions defined on a nestedsequence of regular subspaces Please refer to [14 Lemma 31]and [5] for closely related results

Theorem7 LetU andY be Krein spaces and letH119896 sub U andK119896 sub Y 119896 isin N be sequences of regular subspaces satisfying

H119896 subH119896+1K119896 subK119896+1 (28)

Let 119875H119896 and 119875K119896 be the orthogonal projections of U onto H119896andY ontoK119896 respectively and let H = spanH119896 and K =spanK119896 Assume that there exist fundamental symmetries 119869on U and 1198691015840 on Y such that 119869 commutes with 119875H119896 and 1198691015840commutes with 119875K119896 and that U ⊖ H Y ⊖ K H119896+1 ⊖ H1andK119896+1 ⊖K1 are all Hilbert space

For each 119896 isin N let119860119896 H119896 997888rarrK119896 (29)

be a contraction Then there exists a contraction 119861 U rarr Ysuch that 119861|H119896 = 119860119896 119896 isin N (30)

if and only if 119860119896+11003816100381610038161003816H119896 = 119860119896sup119896isinN

10038171003817100381710038171198601198961003817100381710038171003817 lt infin (31)


Proof First let us assume that such an extension 119861 isin119861(UY) exists and take ℎ119896 isin H119896 Since H119896 sub H119896+1 and(30) holds we have that119860119896+1ℎ119896 = 119861ℎ119896 = 119860119896ℎ119896 (32)

and therefore 119860119896+1|H119896 = 119860119896 Hence the first condition in(31) holds To show that the second condition also holds wefirst note that since 119869 and 1198691015840 commute with 119875H119896 and 119875K119896 respectively Corollary 6 ensures that the subspaces H andK are regular and that there exist fundamental symmetries119869H on H and 1198691015840

Kon K such that 119869H | H119896 is a fundamental

symmetry on H119896 and 1198691015840K | K119896 is a fundamental symmetryonK119896 Let ℎ isinH119896 Then we have that 119860119896ℎ119861ℎ Hence1003817100381710038171003817119860119896ℎ1003817100381710038171003817K = 1003817100381710038171003817119860119896ℎ1003817100381710038171003817K119896 = ⟨1198691015840K119860119896ℎ 119860119896ℎ⟩= ⟨1198691015840

K119861ℎ 119861ℎ⟩ = 119861ℎK = 119861 ℎH (33)

Hence 119860119896 le 119861 for all 119896 isin N and so the norms of 119860119896rsquos areuniformly bounded

Conversely let 119860119896 H119896 rarr K119896 119896 isin N be contractionssatisfying both conditions in (31) Since 119860119896+1|H119896 = 119860119896 wecan decompose the operator

119860119896+11003816100381610038161003816H119896 H119896 997888rarr ( K119896

K119896+1 ⊖K119896) (34)

as 119860119896+1|H119896 = ( 1198601198960 ) The fact that 119860119896 is a contraction andH119896+1 ⊖H119896 is a Hilbert space implies that the operator

(119860119896 0) ( H119896

H119896+1 ⊖H119896) 997888rarrK119896 (35)

is a contraction Since 119860119896+1|H119896 = ( 1198601198960) is a contraction

Theorem 2 implies that there exists a bounded operator 11988822 H119896+1 ⊖H119896 rarrK119896+1 ⊖K119896 such that

119862119896+1 = (119860119896 00 11988822) ( H119896

H119896+1 ⊖H119896) 997888rarr ( K119896

K119896+1 ⊖K119896) (36)

is a contraction For each 119896 = 0 1 2 the operator 119862119896+1 isclearly a contractive extension of 119860119896

We now show that for 119896 = 0 1 2 the contractiveliftings 119862119896+1rsquos of 119860119896rsquos are uniformly bounded Let 119872 be thebound for the norm of 119860119896rsquos Then (119860119896 0) = 119860119896 le 119872and so Lemma 1 implies that 119862119896+12 le 1+21198601198962 le 1+21198722Hence 119862119896+1rsquos are uniformly bounded

Define an operator 1198621015840infin spanH119896 rarr spanK119896 by 1198621015840infin119909 =119862119896+1119909 for 119909 isin H119896 The operator 1198621015840infin is well defined To seethis assume that 119909 isin H119896 and 119909 isin H119897 Then 1198621015840infin119909 = 119862119896+1119909and 1198621015840infin119909 = 119862119897+1119909 For 119896 lt 119897 119862119897+1|H119896 = 119860 119897|H119896 = 119860119896 =119862119896+1|H119896 The operator 1198621015840infin is bounded since the contractions119862119896+1 are uniformly bounded To see that this is the caseconsider 119909 isinH119896 Then[119909 119909]H = ⟨119869H119909 119909⟩H = ⟨119869H119909 119909⟩H119896 = ⟨119869H119896119909 119909⟩H119896= [119909 119909]H119896 (37)

Hence for 119909 isin spanH119896 we see that100381710038171003817100381710038171198621015840infin11990910038171003817100381710038171003817K = 1003817100381710038171003817119862119896+11199091003817100381710038171003817K = 1003817100381710038171003817119862119896+11199091003817100381710038171003817K119896 le 1003817100381710038171003817119862119896+11003817100381710038171003817 119909H119896= 1003817100381710038171003817119862119896+11003817100381710038171003817 119909H (38)

and so 1198621015840infin is a bounded operator Since it is defined ona dense set we can extend it by continuity to a boundedoperator 119862infin H rarr K The operator 119862infin is clearly anextension of 119860119896 for each 119896 ge 0 since 119862119896+1 is an extensionof 119860119896 for each 119896 ge 0 To show that 119862infin is a contraction let119909 isin H Then there exists a sequence 119909119899 isin spanH119896 such that119909119899 rarr 119909 Since the inequality⟨1198621015840infin119909119899 1198621015840infin119909119899⟩ le ⟨119909119899 119909119899⟩ (39)

holds for each 119899 we have⟨119862infin119909 119862infin119909⟩ = lim119899rarrinfin

⟨1198621015840infin119909119899 1198621015840infin119909119899⟩le lim119899rarrinfin

⟨119909119899 119909119899⟩ = ⟨119909 119909⟩ (40)

Hence 119862infin Hrarr K is a contractionDefine a new operator 119861 Urarr Y by the matrix

119861 fl (119862infin 1198640 119883) ( H

U ⊖ H) 997888rarr ( K

Y ⊖ K) (41)

where (119862infin 119864) (42)

is any contractive row extension of 119862infin and 119883 isin 119861((U ⊖H) (Y ⊖ K)) Since Y ⊖ K is a Hilbert space Theorem 2guarantees the existence of 119883 such that 119861 is a contractionNote that since U ⊖ H is also a Hilbert space we may set119864 = 0 the zero operator in the matrix representation of 119861and the above result still holds

Clearly 119861 | H119896 = 119860119896 and so 119861 is the required extension

5 Applications

In this section we use Theorem 7 to solve a nonstationaryextension problem in a Krein space setting Let U119896 andY119895 (119895 119896 isin N) be Krein spaces with fixed fundamentaldecompositions

U119896 = U+119896 oplusUminus119896

Y119895 = Y+119895 oplusYminus119895 (43)

and let 119891119895119896 U119896 997888rarr Y119895 (44)

be bounded operators with matrix representations

119891119895119896 = (11989111119895119896 1198912111989511989611989121119895119896 11989122119895119896) (U+119896

Uminus119896) 997888rarr (Y+119895

Yminus119895) (45)


with11989111119895119896 = 0 for 119895 gt 119896 and11989121119895119896 = 0 for 119895 gt 0 Define the Kreinspaces

997888rarrU and

997888rarrY by

997888rarrU fl U0 oplusU1 oplusU2 oplus sdot sdot sdot 997888rarrY fl Y0 oplusY1 oplusY2 oplus sdot sdot sdot (46)

and let 119866 997888rarrU to997888rarrY be a bounded operator such that

119875Y119895119866100381610038161003816100381610038161003816U119896 = 119891119895119896 (47)

We use Theorem 7 to establish conditions under which suchan operator 119866 is a contraction Equality (47) implies that theoperator 119866 is of the form

119866 =(((((((

sdot sdot sdot 11989140 11989130 11989120 11989110 11989100sdot sdot sdot 11989141 11989131 11989121 11989111 11989101sdot sdot sdot 11989142 11989132 11989122 11989112 11989102sdot sdot sdot 11989143 11989133 11989123 11989113 11989103c

)))))))

(((((((((((

U4

U3

U2

U1

U0

)))))))))))

997888rarr(((((((((((

Y0

Y1

Y2

Y3

Y4

)))))))))))

(48)

Consider the decompositions

997888rarrU = sdot sdot sdot oplusUminus1 oplusUminus0 oplusU+0 oplusU+1 oplus sdot sdot sdot 997888rarrY = sdot sdot sdot oplusYminus1 oplusYminus0 oplusY+0 oplusY+1 oplus sdot sdot sdot (49)

With the above decompositions the operator 119866 takes theform

119866 =((((((((((((((((

d csdot sdot sdot 1198911123 1198911122 0 0 0 0 0 0 sdot sdot sdotsdot sdot sdot 1198911113 1198911112 1198911111 0 0 0 0 0 sdot sdot sdot

sdot sdot sdot 1198911103 1198911102 1198911101 1198911100 1198911200 1198911201 1198911202 1198911203 sdot sdot sdotsdot sdot sdot 1198912103 1198912102 1198912101 1198912100 1198912200 1198912201 1198912202 1198912203 sdot sdot sdotsdot sdot sdot 1198911113 1198911112 1198911111 1198911110 1198912210 1198912211 1198912212 1198912213 sdot sdot sdotc

))))))))))))))))

((((((((((((((((((

U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))))

997888rarr((((((((((((((((((

Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))))

(50)

For each 119896 isin N let 119860119896 denote the operator119860119896

=((((((((((((((((

11989111119896119896 0 sdot sdot sdot 0 0 0 0 0 0 sdot sdot sdot d 119891112119896 sdot sdot sdot 1198911122 0 0 0 0 0 0 sdot sdot sdot119891111119896 sdot sdot sdot 1198911112 1198911111 0 0 0 0 0 sdot sdot sdot119891110119896 sdot sdot sdot 1198911102 1198911101 1198911100 1198911200 1198911201 1198911202 1198911203 sdot sdot sdot119891210119896 sdot sdot sdot 1198912102 1198912101 1198912100 1198912200 1198912201 1198912202 1198912203 sdot sdot sdot119891211119896 sdot sdot sdot 1198912112 1198912111 1198912110 1198912210 1198912211 1198912212 1198912213 sdot sdot sdot

))))))))))))))))

((((((((((((((((

U119896U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))

997888rarr((((((((((((((((

Y119896Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))

(51)

With these notations we are in a position to state thefollowing theorem

Theorem 8 Let U119896 and Y119895 be Krein spaces with decomposi-tions (43) with operators 119891119895119896 U119896 rarr Y119895 in (44) Then thereexists a contraction 119866 997888rarrU rarr 997888rarr

Y satisfying (47) if and only iffor each 119896 isin N the operator 119860119896 in (51) defines a contractionwhose norm has an upper bound independent of 119896


Proof Assume that a contraction 119866 997888rarrUrarr 997888rarrY satisfying (47)

exists and consider the operator matrices (50) and (51) Wesee that for 119896 isin N 119860119896 = 119875K11989611986610038161003816100381610038161003816H119896 (52)

where

H119896 =((((((((((

U119896U+1

U+0Uminus0

Uminus1

))))))))))

K119896 =((((((((((

Y119896Y+1

Y+0Yminus0

Yminus1

))))))))))

(53)

Since997888rarrY ⊖K119896 is a Hilbert space we see that 119875K119896 le 1 and

so 10038171003817100381710038171198601198961003817100381710038171003817 le 10038171003817100381710038171003817119875K11989610038171003817100381710038171003817 119866 le 119866 (54)

This shows that the operators 119860119896 are uniformly boundedSince 119866 is a contraction and

997888rarrY ⊖K119896 is a Hilbert space (52)

implies that 119860119896 is a contraction for each 119896 isin NTo prove the reverse implication we assume that for each119896 isin N the operator 119860119896 H119896 rarr K119896 is a contraction of

norm at most 120574 where 0 lt 120574 lt infin Since spanH119896 = 997888rarrUand spanK119896 = 997888rarrY it follows that spanH119896 and spanK119896 areregular subspaces By construction the subspaces H119896 andK119896 satisfy the conditions of Corollary 6 and hence do satisfyall the conditions specified in Theorem 7 Since for each119896 isin N the operators 119860119896 H119896 997888rarrK119896 (55)

are defined by (51) and are assumed to be uniformly boundedcondition (31) in Theorem 7 is also fulfilled (it can be easilyseen that the first condition in (31) is satisfied by the operators119860119896) Since (31) is satisfied for all 119896 isin N it follows that thereexists a contraction 119861 satisfying (30) and therefore must be ofthe form (50) This concludes the proof

Conflicts of Interest

The author declares that there are no conflicts of interestregarding the publication of this paper

References

[1] G Arsene T Constantinescu and A Gheondea ldquoLifting ofoperators and prescribed numbers of negative squaresrdquo Michi-gan Mathematical Journal vol 34 no 2 pp 201ndash216 1987

[2] M A Dritschel and J Rovnyak ldquoExtension theorems for con-traction operators on Krein spacesrdquo Operator Theory Advancesand Applications vol 47 pp 221ndash305 1990

[3] M A Dritschel ldquoA lifting theorem for bicontractions on Kreınspacesrdquo Journal of Functional Analysis vol 89 no 1 pp 61ndash891990

[4] A Gheondea ldquoCanonical forms of unbouded unitary operatorsin Krein operatorsrdquo Publications of the Research Institute forMathematical Sciences vol 24 no 2 pp 205ndash224 1988

[5] A Gheondea ldquoOn the structure of linear contractions withrespect to fundamental decompositionsrdquo Operator TheoryAdvances and Applications vol 48 pp 275ndash295 1990

[6] C Foias A E Frazho I Gohberg and M Kaashoek ldquoMetricconstrained interpolation commutant lifting and systemsrdquoOperator Theory Advances and Applications vol 100 1998

[7] C Foias A E Frazho I Gohberg and M Kaashoek ldquoParam-eterization of all solutions of the three chains completionproblemrdquo Integral Equations and Operator Theory vol 29 no4 pp 455ndash490 1997

[8] C Foias A E Frazho I Gohberg and M Kaashoek ldquoAtime-variant version of the commutant lifting theorem andnonstationary interpolation problemsrdquo Integral Equations andOperator Theory vol 28 no 2 pp 158ndash190 1997

[9] T Ando Linear Operators on Krein Spaces Hokkaido Uni-versity Research Institute of Applied Electricity Division ofApplied Mathematics Sapporo Japan 1979

[10] T J Azizov and I S Iohvidov ldquoLinear operators in spaces withan indefinite metricrdquo Itogi Nauki i Techniki pp 113ndash205 2721979

[11] J Bognar Indefinite Inner Product Spaces Springer-Verlag NewYork NY USA 1974

[12] I S Iohvidov M G Krein and H Langer Introduction toThe Spectral Theory of Operators in Spaces with An IndefiniteMetric vol 9 of Mathematical Research Akademie-VerlagBerlin Germany 1982

[13] G Arsene andA Gheondea ldquoCompletingmatrix contractionsrdquoThe Journal of Operator Theory vol 7 no 1 pp 179ndash189 1982

[14] T Constantinescu and A Gheondea ldquoMinimal signature inlifting of operators Irdquo The Journal of Operator Theory vol 22pp 345ndash367 1989

Hindawiwwwhindawicom Volume 2018

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International Journal of


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Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences


Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018


Decision SciencesAdvances in


AnalysisInternational Journal of


Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom


on them This is done in Section 2 Some important resultsare stated and proved in Section 3 An extension theorem isconsidered in Section 4 while Section 5 contains some simpleapplication of the extension theorem discussed in Section 4

2 Preliminaries

In this section we recall some definitions and basic notions ofindefinite metric space theory More details can be obtainedfrom [2 9ndash12]

LetK be a linear space and let ⟨sdot sdot⟩ be a sesquilinear formonKThis sesquilinear form is called an indefinite metric onK IfK admits a direct orthogonal sum decomposition

K =K+ oplusKminus (5)

in which (Kplusmn plusmn⟨sdot sdot⟩) are Hilbert spaces then K (or(K ⟨sdot sdot⟩)) is called aKrein space Decomposition (5) which isnot unique in general is called a fundamental decompositionand gives rise to orthogonal projections from K onto Kplusmnwhich we denote by 119875plusmn respectively The self-adjoint andunitary operator 119869 on K defined by 119869 = 119875+ minus 119875minus is calleda fundamental symmetry onK

The spaceK with the inner product[119909 119910] = ⟨119869119909 119910⟩ = ⟨119909+ 119910+⟩ minus ⟨119909minus 119910minus⟩ 119909plusmn 119910plusmn isinKplusmn (6)

is a Hilbert space This inner product is used to define thenorm of an element 119909 of the Krein spaceK by1199092 = [119909 119909] (7)

Topological notions such as convergence and continuity areunderstood to be with respect to this norm topology Theinner product [sdot sdot] in (6) depends on decomposition (5) asdoes the norm sdot in (7) but the norms generated by differentdecompositions ofK are equivalentWewill denote the norm sdot in (7) by sdot 119869 where clarity is needed

An orthogonal projection in a Krein space is a boundedself-adjoint operator 119877 inK such that 1198772 = 119877 Note that thenorm of an orthogonal projection in a Krein space need notbe less than 1 The rangeM of an orthogonal projection 119877 ina Krein space K is a closed subspace of K and the space Kcan be decomposed as

K =M oplusMperp (8)

where Mperp = 119897 isin K ⟨119897 119898⟩ = 0 forall119898 isin MOn the other hand given a closed subspace M of Ksuch that decomposition (8) holds then M is the range ofan orthogonal projection in K In this case (M ⟨sdot sdot⟩) isagain a Krein space Unlike in the Hilbert space case thedecomposition (8) need not hold for a given closed subspaceClosed subspaces for which this decomposition holds arereferred to as regular subspaces

Let H and K be Krein spaces and let 119879 H rarr K be abounded linear operator We say that 119879 is a contraction if forall 119909 isinH ⟨119879119909 119879119909⟩K le ⟨119909 119909⟩H If both119879 and its Krein spaceadjoint 119879lowast are contractions then 119879 is called a bicontraction

Let H and K be Krein spaces and let 119879 isin 119861(HK) thespace of bounded linear operators fromH intoK A columnextension of 119879 is an operator of the form

119862 = (119879119864) H 997888rarr (KE) (9)

whereE is aKrein space and119864 isin 119861(HE) By a row extensionof 119879 we shall mean an operator of the form

119871 = (119879 119864) (HE) 997888rarrK (10)

where E is a Krein space and 119864 isin 119861(EK) It is shownin [2] (see also [3]) that if 119879 is a contraction then thereexist contractive row and column extensions of 119879 where theextension space is a Hilbert space Contractive 2 times 2 matrixextensions of a contractive operator 119879 H rarr K where theextending spaceE is a Hilbert space are thoroughly discussedin [2]where Lemma 1Theorem2 andLemma3 can be foundResults more general than those provided by Lemma 1 andTheorem 2 can be found in [1] while Lemma 3 can also befound in [4]

Lemma 1 LetH andK be Krein spaces and let119879 isin 119861(HK)be a contraction Let

119862 = ( 119864lowast) isin 119861 (HK oplus E) (11)

be a contractive column extension of 119879 with E a Hilbert spaceIf norms are computed relative to some fixed fundamentaldecompositions of H and K and the induced fundamentaldecomposition ofK oplus E then1198622 le 1 + 2 1198792 (12)

The above norm estimate enables one to fix a bound forthe norm of the operator 119862 The following lemma is helpfulin finding 2times2matrix extensions of a contractive operator 119879See [13] for a similar result in a Hilbert space setting

Theorem 2 LetHK and E be Krein spaces and let E be aHilbert space Assume that

(11986211 11986212) (HE) 997888rarrK

(1198621111986221) H 997888rarr (KE) (13)

are contractions Then there exists an operator 11986222 E rarr Esuch that

119862 = (11986211 1198621211986221 11986222) (HE) 997888rarr (KE) (14)

is a contraction If (11986211 11986212) (15)

is a bicontraction11986222may be chosen such that119862 is a bicontrac-tion




N = infin⋂119899=1

N119899M = spanNperp119899 119899 = 1 2

(16)







L = spanL119899 (17)










= ⟨119869119899119909119899 119909119899⟩ + ⟨119869119899119910119899 119910119899⟩ + ⟨119869perp119911119899 119911119899⟩ (20)


(21)













119875119899 = (11986011989911 1198601198991211986011989921 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (24)


(1 00 minus1)(11986011989911 1198601198991211986011989921 11986011989922) = (119860

11989911 1198601198991211986011989921 11986011989922)(1 00 minus1) (25)


119875119899 = (11986011989911 00 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (26)


L119899+1 =L+119899+1 oplusLminus119899+1 (27)





H119896 subH119896+1K119896 subK119896+1 (28)





10038171003817100381710038171198601198961003817100381710038171003817 lt infin (31)






K119861ℎ 119861ℎ⟩ = 119861ℎK = 119861 ℎH (33)



119860119896+11003816100381610038161003816H119896 H119896 997888rarr ( K119896

K119896+1 ⊖K119896) (34)


(119860119896 0) ( H119896

H119896+1 ⊖H119896) 997888rarrK119896 (35)



119862119896+1 = (119860119896 00 11988822) ( H119896

H119896+1 ⊖H119896) 997888rarr ( K119896

K119896+1 ⊖K119896) (36)








⟨119909119899 119909119899⟩ = ⟨119909 119909⟩ (40)


119861 fl (119862infin 1198640 119883) ( H

U ⊖ H) 997888rarr ( K

Y ⊖ K) (41)

where (119862infin 119864) (42)



5 Applications


U119896 = U+119896 oplusUminus119896

Y119895 = Y+119895 oplusYminus119895 (43)

and let 119891119895119896 U119896 997888rarr Y119895 (44)


119891119895119896 = (11989111119895119896 1198912111989511989611989121119895119896 11989122119895119896) (U+119896

Uminus119896) 997888rarr (Y+119895

Yminus119895) (45)



997888rarrU and

997888rarrY by



119875Y119895119866100381610038161003816100381610038161003816U119896 = 119891119895119896 (47)


119866 =(((((((


)))))))

(((((((((((

U4

U3

U2

U1

U0

)))))))))))

997888rarr(((((((((((

Y0

Y1

Y2

Y3

Y4

)))))))))))

(48)




119866 =((((((((((((((((



))))))))))))))))

((((((((((((((((((

U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))))

997888rarr((((((((((((((((((

Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))))

(50)


=((((((((((((((((


))))))))))))))))

((((((((((((((((

U119896U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))

997888rarr((((((((((((((((

Y119896Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))

(51)







where

H119896 =((((((((((

U119896U+1

U+0Uminus0

Uminus1

))))))))))

K119896 =((((((((((

Y119896Y+1

Y+0Yminus0

Yminus1

))))))))))

(53)


so 10038171003817100381710038171198601198961003817100381710038171003817 le 10038171003817100381710038171003817119875K11989610038171003817100381710038171003817 119866 le 119866 (54)








References






















Journal of












Journal of







Volume 2018






Volume 2018











N = infin⋂119899=1

N119899M = spanNperp119899 119899 = 1 2

(16)







L = spanL119899 (17)










= ⟨119869119899119909119899 119909119899⟩ + ⟨119869119899119910119899 119910119899⟩ + ⟨119869perp119911119899 119911119899⟩ (20)


(21)













119875119899 = (11986011989911 1198601198991211986011989921 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (24)


(1 00 minus1)(11986011989911 1198601198991211986011989921 11986011989922) = (119860

11989911 1198601198991211986011989921 11986011989922)(1 00 minus1) (25)


119875119899 = (11986011989911 00 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (26)


L119899+1 =L+119899+1 oplusLminus119899+1 (27)





H119896 subH119896+1K119896 subK119896+1 (28)





10038171003817100381710038171198601198961003817100381710038171003817 lt infin (31)






K119861ℎ 119861ℎ⟩ = 119861ℎK = 119861 ℎH (33)



119860119896+11003816100381610038161003816H119896 H119896 997888rarr ( K119896

K119896+1 ⊖K119896) (34)


(119860119896 0) ( H119896

H119896+1 ⊖H119896) 997888rarrK119896 (35)



119862119896+1 = (119860119896 00 11988822) ( H119896

H119896+1 ⊖H119896) 997888rarr ( K119896

K119896+1 ⊖K119896) (36)








⟨119909119899 119909119899⟩ = ⟨119909 119909⟩ (40)


119861 fl (119862infin 1198640 119883) ( H

U ⊖ H) 997888rarr ( K

Y ⊖ K) (41)

where (119862infin 119864) (42)



5 Applications


U119896 = U+119896 oplusUminus119896

Y119895 = Y+119895 oplusYminus119895 (43)

and let 119891119895119896 U119896 997888rarr Y119895 (44)


119891119895119896 = (11989111119895119896 1198912111989511989611989121119895119896 11989122119895119896) (U+119896

Uminus119896) 997888rarr (Y+119895

Yminus119895) (45)



997888rarrU and

997888rarrY by



119875Y119895119866100381610038161003816100381610038161003816U119896 = 119891119895119896 (47)


119866 =(((((((


)))))))

(((((((((((

U4

U3

U2

U1

U0

)))))))))))

997888rarr(((((((((((

Y0

Y1

Y2

Y3

Y4

)))))))))))

(48)




119866 =((((((((((((((((



))))))))))))))))

((((((((((((((((((

U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))))

997888rarr((((((((((((((((((

Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))))

(50)


=((((((((((((((((


))))))))))))))))

((((((((((((((((

U119896U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))

997888rarr((((((((((((((((

Y119896Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))

(51)







where

H119896 =((((((((((

U119896U+1

U+0Uminus0

Uminus1

))))))))))

K119896 =((((((((((

Y119896Y+1

Y+0Yminus0

Yminus1

))))))))))

(53)


so 10038171003817100381710038171198601198961003817100381710038171003817 le 10038171003817100381710038171003817119875K11989610038171003817100381710038171003817 119866 le 119866 (54)








References






















Journal of












Journal of







Volume 2018






Volume 2018
















119875119899 = (11986011989911 1198601198991211986011989921 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (24)


(1 00 minus1)(11986011989911 1198601198991211986011989921 11986011989922) = (119860

11989911 1198601198991211986011989921 11986011989922)(1 00 minus1) (25)


119875119899 = (11986011989911 00 11986011989922) (K+Kminus

) 997888rarr (L+119899Lminus119899

) (26)


L119899+1 =L+119899+1 oplusLminus119899+1 (27)





H119896 subH119896+1K119896 subK119896+1 (28)





10038171003817100381710038171198601198961003817100381710038171003817 lt infin (31)






K119861ℎ 119861ℎ⟩ = 119861ℎK = 119861 ℎH (33)



119860119896+11003816100381610038161003816H119896 H119896 997888rarr ( K119896

K119896+1 ⊖K119896) (34)


(119860119896 0) ( H119896

H119896+1 ⊖H119896) 997888rarrK119896 (35)



119862119896+1 = (119860119896 00 11988822) ( H119896

H119896+1 ⊖H119896) 997888rarr ( K119896

K119896+1 ⊖K119896) (36)








⟨119909119899 119909119899⟩ = ⟨119909 119909⟩ (40)


119861 fl (119862infin 1198640 119883) ( H

U ⊖ H) 997888rarr ( K

Y ⊖ K) (41)

where (119862infin 119864) (42)



5 Applications


U119896 = U+119896 oplusUminus119896

Y119895 = Y+119895 oplusYminus119895 (43)

and let 119891119895119896 U119896 997888rarr Y119895 (44)


119891119895119896 = (11989111119895119896 1198912111989511989611989121119895119896 11989122119895119896) (U+119896

Uminus119896) 997888rarr (Y+119895

Yminus119895) (45)



997888rarrU and

997888rarrY by



119875Y119895119866100381610038161003816100381610038161003816U119896 = 119891119895119896 (47)


119866 =(((((((


)))))))

(((((((((((

U4

U3

U2

U1

U0

)))))))))))

997888rarr(((((((((((

Y0

Y1

Y2

Y3

Y4

)))))))))))

(48)




119866 =((((((((((((((((



))))))))))))))))

((((((((((((((((((

U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))))

997888rarr((((((((((((((((((

Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))))

(50)


=((((((((((((((((


))))))))))))))))

((((((((((((((((

U119896U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))

997888rarr((((((((((((((((

Y119896Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))

(51)







where

H119896 =((((((((((

U119896U+1

U+0Uminus0

Uminus1

))))))))))

K119896 =((((((((((

Y119896Y+1

Y+0Yminus0

Yminus1

))))))))))

(53)


so 10038171003817100381710038171198601198961003817100381710038171003817 le 10038171003817100381710038171003817119875K11989610038171003817100381710038171003817 119866 le 119866 (54)








References






















Journal of












Journal of







Volume 2018






Volume 2018













K119861ℎ 119861ℎ⟩ = 119861ℎK = 119861 ℎH (33)



119860119896+11003816100381610038161003816H119896 H119896 997888rarr ( K119896

K119896+1 ⊖K119896) (34)


(119860119896 0) ( H119896

H119896+1 ⊖H119896) 997888rarrK119896 (35)



119862119896+1 = (119860119896 00 11988822) ( H119896

H119896+1 ⊖H119896) 997888rarr ( K119896

K119896+1 ⊖K119896) (36)








⟨119909119899 119909119899⟩ = ⟨119909 119909⟩ (40)


119861 fl (119862infin 1198640 119883) ( H

U ⊖ H) 997888rarr ( K

Y ⊖ K) (41)

where (119862infin 119864) (42)



5 Applications


U119896 = U+119896 oplusUminus119896

Y119895 = Y+119895 oplusYminus119895 (43)

and let 119891119895119896 U119896 997888rarr Y119895 (44)


119891119895119896 = (11989111119895119896 1198912111989511989611989121119895119896 11989122119895119896) (U+119896

Uminus119896) 997888rarr (Y+119895

Yminus119895) (45)



997888rarrU and

997888rarrY by



119875Y119895119866100381610038161003816100381610038161003816U119896 = 119891119895119896 (47)


119866 =(((((((


)))))))

(((((((((((

U4

U3

U2

U1

U0

)))))))))))

997888rarr(((((((((((

Y0

Y1

Y2

Y3

Y4

)))))))))))

(48)




119866 =((((((((((((((((



))))))))))))))))

((((((((((((((((((

U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))))

997888rarr((((((((((((((((((

Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))))

(50)


=((((((((((((((((


))))))))))))))))

((((((((((((((((

U119896U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))

997888rarr((((((((((((((((

Y119896Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))

(51)







where

H119896 =((((((((((

U119896U+1

U+0Uminus0

Uminus1

))))))))))

K119896 =((((((((((

Y119896Y+1

Y+0Yminus0

Yminus1

))))))))))

(53)


so 10038171003817100381710038171198601198961003817100381710038171003817 le 10038171003817100381710038171003817119875K11989610038171003817100381710038171003817 119866 le 119866 (54)








References






















Journal of












Journal of







Volume 2018






Volume 2018










997888rarrU and

997888rarrY by



119875Y119895119866100381610038161003816100381610038161003816U119896 = 119891119895119896 (47)


119866 =(((((((


)))))))

(((((((((((

U4

U3

U2

U1

U0

)))))))))))

997888rarr(((((((((((

Y0

Y1

Y2

Y3

Y4

)))))))))))

(48)




119866 =((((((((((((((((



))))))))))))))))

((((((((((((((((((

U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))))

997888rarr((((((((((((((((((

Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))))

(50)


=((((((((((((((((


))))))))))))))))

((((((((((((((((

U119896U+2

U+1

U+0Uminus0

Uminus1

Uminus2

))))))))))))))))

997888rarr((((((((((((((((

Y119896Y+2

Y+1

Y+0Yminus0

Yminus1

Yminus2

))))))))))))))))

(51)







where

H119896 =((((((((((

U119896U+1

U+0Uminus0

Uminus1

))))))))))

K119896 =((((((((((

Y119896Y+1

Y+0Yminus0

Yminus1

))))))))))

(53)


so 10038171003817100381710038171198601198961003817100381710038171003817 le 10038171003817100381710038171003817119875K11989610038171003817100381710038171003817 119866 le 119866 (54)








References






















Journal of












Journal of







Volume 2018






Volume 2018











where

H119896 =((((((((((

U119896U+1

U+0Uminus0

Uminus1

))))))))))

K119896 =((((((((((

Y119896Y+1

Y+0Yminus0

Yminus1

))))))))))

(53)


so 10038171003817100381710038171198601198961003817100381710038171003817 le 10038171003817100381710038171003817119875K11989610038171003817100381710038171003817 119866 le 119866 (54)








References






















Journal of












Journal of







Volume 2018






Volume 2018















Journal of












Journal of







Volume 2018






Volume 2018








an extension theorem for a sequence of krein space...

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